I need help with the following series: $$ S(x)=\sum _{n=1}^{+\infty} \frac{(-1)^n \hspace{0.05cm} I_{n}^2 (x)}{2n-1}, \hspace{0.3cm}x \in \mathbb{R}_{>0} $$ where $I_n(x)$ is the modified Bessel function of first kind. Is it possible to obtain a nice relation for it (i.e. not in the form of an infinite sum)?
There is this formula (eq. 2.3 in the M. D. Rogers paper): $$ \sum _{n=-\infty }^{\infty } \frac{J_n^2(z)}{n+y}=J_{y}(z) J_{-y}(z)\frac{\pi}{\sin(\pi y)} $$ If we set $y=-\frac{1}{2}$, $z=ix$ and use $J_n(ix)=i^nI_n(x)$, $J_{-1/2}(z)=\left(\frac2{\pi z}\right)^{1/2}\cos z$, $J_{1/2}(z)=\left(\frac2{\pi z}\right)^{1/2}\sin z$, and $\sin(2ix)=i\sinh(2x)$ , we get $$ \sum _{n=-\infty}^{+\infty} \frac{(-1)^n \hspace{0.05cm} I_{n}^2 (x)}{2n-1}=-\frac{\sinh(2x)}{2x} $$ But it's not the sum $S(x)$, which starts from $n=1$ and not $n=-\infty$.
I don't know if it helps, but there is also a formula (from "Integrals and Series Vol. 2: Special Functions", also here a similar formula) $$ \sum _{n=1}^{\infty } \frac{(-1)^n I_{n\nu}^2 (z)}{n^2-a^2}=\frac{I_{\nu}^2(z)}{2 a^2}-\frac{\pi \csc (\pi a) I_{\nu a}^2(z)}{2 a}. $$ but I am not sure if we can set $\nu=1$ (but even if we can, it's still not very helpful).